The inclination (tilt) of an amusement park ride is accelerating at a rate of $2160\,\dfrac{\text{degrees}}{\text{min}^2}$. What is the ride's acceleration rate in $\dfrac{\text{degrees}}{\text{s}^2}$ ?
Answer: We will convert $2160\,\dfrac{\text{degrees}}{\text{min}^2}$ to an acceleration rate in $\dfrac{\text{degrees}}{\text{s}^2}$ using the following conversion rate: There are $60\text{ s}$ per $1\text{ min}$. Therefore there are $(60\text{ s})^2=3600\text{ s}^2$ per $1\text{ min}^2$. $\begin{aligned} &\phantom{=} \dfrac{2160 \text{ degrees}}{1\text{ min}^2} \cdot\dfrac{1\text{ min}^2}{3600\text{ s}^2} \\\\ &=\dfrac{2160 \cdot 1 \cdot\text{degrees} \cdot \cancel{\text{min}^2}}{1\cdot 3600 \cdot\cancel{\text{min}^2} \cdot \text{s}^2 } \\\\ &=\dfrac{2160}{3600}\,\dfrac{\text{degrees}}{\text{s}^2} \\\\ &=\dfrac{3}{5}\,\dfrac{\text{degrees}}{\text{s}^2} \end{aligned}$ In conclusion, the acceleration rate in $\dfrac{\text{degrees}}{\text{s}^2}$ is: $\dfrac{3}{5}\,\dfrac{\text{degrees}}{\text{s}^2}$